Math 161
Homework 3 Solutions
Summer 2015
Drew Armstrong
Book Problems:
•
•
•
•
Chap
Chap
Chap
Chap
2.5
2.8
3.3
3.5
Exercises
Exercises
Exercises
Exercises
8, 22, 24, 38
11, 12, 22, 24
22, 34
2, 4
Solutions:
2.5.8. Let y = (4x − x2 )100 . Compute
dy
dx .
We’ll use the chain rule. Let u = 4x − x2 so that y = u100 . Then
dy
dy du
=
·
dx
du dx
= 100 u99 · (4 − 2x)
= 100 (4x − x2 )99 (4 − 2x).
r
2.5.22. Let f (s) =
s2 + 1
. Compute f 0 (s).
s2 + 4
We’ll use the chain rule, and this time we won’t be so fussy about it.
− 21 2
0
1 s2 + 1
s +1
0
f (s) =
·
2 s2 + 4
s2 + 4
1
− 2
1 s2 + 1
(s2 + 4)(s2 + 1)0 − (s2 + 1)(s2 + 4)0
·
=
2 s2 + 4
(s2 + 4)2
2
− 21
1 s +1
(s2 + 4)(2s) − (s2 + 1)(2s)
=
·
2 s2 + 4
(s2 + 4)2
− 21
1 s2 + 1
2s((s2 + 4) − (s2 + 1))
·
=
2 s2 + 4
(s2 + 4)2
− 21
1 s2 + 1
6s
· 2
=
.
2
2 s +4
(s + 4)2
We won’t bother to simplify it further. (I don’t even know why I even simplified it this far.)
2.5.24. Let f (x) = √
x
. Compute f 0 (x).
7 − 3x
I’ll use the quotient rule:
√
0
f (x) =
√
=
√
=
√
7 − 3x · (x)0 − x · ( 7 − 3x)0
√
( 7 − 3x)2
1
7 − 3x · (1) − x · ( 2√7−3x
· (−3))
7 − 3x
7 − 3x +
√3x
2 7−3x
7 − 3x
We don’t√need to simplify, but if we want to we can multiply the numerator and denominator
both by 7 − 3x to get
√
√
7 − 3x + 2√3x
7 − 3x
7−3x
0
f (x) =
·√
7 − 3x
7 − 3x
3x
(7 − 3x) + 2
√
=
(7 − 3x) 7 − 3x
7 − 3x
2
(7 − 3x)3/2
14 − 3x
=
.
2(7 − 3x)3/2
=
q
p
√
2.5.38. Let y = x + x + x. Compute
dy
dx .
Just do the chain rule a lot:
1
q
q
√ −2 d
√
x+ x+ x
·
x+ x+ x
dx
1 q
√ −2
√ − 12 d
√ 1
1
· 1+
=
x+ x+ x
x+ x
·
x+ x
2
2
dx
1 q
√ −2
√ − 12
1
1
1 −1
=
· 1+ x 2
x+ x+ x
· 1+
x+ x
2
2
2
dy
1
=
dx
2
Don’t bother simplifying. There’s no way to make this look good.
2.8.11. Use linear approximation to estimate (1.999)4 .
We want to estimate the value of f (x) = x4 for x near 2. The formula is
f (x) ≈ f (2) + f 0 (2)(x − 2) when x ≈ 2.
Since f 0 (x) = 4x3 , the boxed formula says
x4 ≈ (2)4 + 4(2)3 (x − 2)
x4 ≈ 16 + 32(x − 2)
x4 ≈ −48 + 32x.
Of course, this is only accuate when x ≈ 2. Luckily, 1.999 ≈ 2 so we get
(1.999)4 ≈ −48 + 32(1.999) = 15.968
How good is this estimate? My computer says (1.999)4 = 15.96802, so it’s pretty good.
√
2.8.12. Use linear approximation to estimate 3 1001 = (1001)1/3 .
We want to estimate the value of f (x) = x1/3 for x near 1000. The formula is
f (x) ≈ f (1000) + f 0 (1000)(x − 1000) when x ≈ 1000.
2
Since f 0 (x) = 31 x− 3 , the boxed formula says
2
1
x1/3 ≈ (1000)1/3 + (1000)− 3 (x − 1000)
3
1 1
1/3
(x − 1000)
x ≈ 10 + ·
3 100
20
x
x1/3 ≈
+
3
300
Of course, this is only accurate when x ≈ 1000. Luckily, 1001 ≈ 1000 so we get
20 1001
3001
(1001)1/3 ≈
−
=
= 10.00333 · · · .
3
300
300
How good is this estimate? My computer says (1001)1/3 = 10.0033322, so it’s pretty good.
2.8.22. The radius of a circular disk is given as 24 cm with a maximum error in the measurement of 0.2 cm. Use differentials to estimate the maximum error in the area of the disk.
What is the relative error? What is the percentage error?
Let r stand for the radius of the disk, so that r = 24 and dr = 0.2. Let A stand for the
area of the disk. Then we have A = πr2 . We want to compute the differential dA. Well,
dA
= 2πr
dr
dA = 2πr dr
so we have dA = 2π(24)(0.2) = 30.16 cm2 . This is our estimate for the maximum error in the
area of the disk. To compute the relative error we first need to know that A = πr2 = π(24)2 =
1809.56 cm2 . Thus the relative (and percentage) error is
dA
30.16
=
= 0.017 = 1.7%.
A
1809.56
2.8.24. Use differentials to estimate the amount of paint needed to apply a coat of paint 0.05
cm thick to a hemispherical dome with diameter 50 m.
This is a fun one. Let D represent the diameter of the dome, so D = 50 m. When we add
the coat of paint (I’m assuming on the outside of the dome, but it could be on the inside I
guess), the diameter of the dome will change by dD = 2(0.05) = 0.1 cm.
Now we want to know how much paint is needed. Paint is measured by volume, so let
V represent volume. Of what? This is tricky. We let V represent the volume of the full
hemisphere. Then dV is the change in the volume of the hemisphere after we add the paint.
In other words, dV is the volume of paint!
The volume of a hemisphere is 1/2 the volume of a sphere:
1 4
V = · π
2 3
D
2
3
=
π 3
D .
12
The differential is
π
π
dV
=
· 3D2 = D2
dD
12
4
π 2
dV = D dD
4
Thus the amount of paint required is
dV =
π
(50)2 (0.001) = 1.963 m3 .
4
(Note that I rewrote 0.1 cm as 0.001 m.) In summary, we will need 1.963 cubic meters of paint
to cover the dome. Google tells me that this is 518.57 US gallons. Is that more or less than
you expected?
3.3.22. Let f (x) = 36x + 3x2 − 2x3 . Determine when f 0 (x) and f 00 (x) are positive, negative,
and zero. Use this information to sketch the graph.
First we compute
f 0 (x) = 36 + 3 · 2x − 2 · 3x2
= 36 + 6x − 6x2
= 6(6 + x − x2 )
= 6(2 + x)(3 − x).
We conclude that f 0 (x) = 0 when x = −2 or x = 3. These are called the “critical numbers”.
In between we have f 0 (x) < 0 (f is decreasing) when x < −2, f 0 (x) > 0 (f is increasing) when
−2 < x < 3, and f 0 (x) < 0 (f is decreasing) when 3 < x.
Next we compute
f 00 (x) = (36 + 6x − 6x2 )0
= 0 + 6 − 6 · 2x
= 6 − 12x
= 6(1 − 2x).
We conclude that f 00 (x) = 0 when x = 1/2 (this is an inflection point). When x < 1/2 we
have f 00 (x) > 0 (f is concave up), and when 1/2 < x we have f 00 (x) < 0 (f is concave down).
Evaluating f 00 (x) at the critical numbers gives f 00 (−2) > 0 and f 00 (3) < 0, so by the second
derivative test we conclude that f (x) has a local minimum at x = −2 and a local maximum
at x = 3.
In summary, the graph of f (x) has a minimum at (−2, f (−2)) = (−2, −44), an inflection
at (1/2, f (1/2)) = (1/2, 37/2) and a maximum at (3, f (3)) = (3, 81). Here is a sketch showing
the graph of f (x), the extrema, and the inflection points, made by my computer:
3.3.34. Let f (x) = (x2 − 4)/(x2 + 4). Determine when f 0 (x) and f 00 (x) are positive, negative,
and zero. Use this information to sketch the graph.
First we use the quotient rule to compute
(x2 + 4)(x2 − 4)0 − (x2 − 4)(x2 + 4)0
(x2 + 4)2
(x2 + 4)(2x) − (x2 − 4)(2x)
=
(x2 + 4)2
2x((x2 + 4) − (x2 − 4))
=
(x2 + 4)2
16x
= 2
(x + 4)2
f 0 (x) =
Since (x2 + 4)2 is always strictly positive, we conclude that: f 0 (x) < 0 when x < 0, f 0 (x) = 0
when x = 0, and f 0 (x) > 0 when x > 0. The only critical number is x = 0.
Next we use the quotient rule again to compute
f 00 (x) =
=
=
=
=
(x2 + 4)2 (16x)0 − 16x((x2 + 4)2 )0
[(x2 + 4)2 ]2
16(x2 + 4)2 − 16x · 2(x2 + 4)(2x)
(x2 + 4)4
2
16(x + 4)[(x2 + 4) − 4x2 ]
(x2 + 4)4
16(x2 + 4)(4 − 3x2 )
(x2 + 4)4
16(4 − 3x2 )
.
(x2 + 4)3
3
Since (x2 + 4)
positive, we conclude
that f 00 (x) = 0 when 4 − 3x2 = 0, i.e.,
p
p
p is always strictly
when x = + 4/3 or x = − 4/3. When x < − 4/3 we have f 00 (x) < 0 (f is concave down),
p
p
p
when − 4/3 < x < + 4/3 we have f 00 (x) > 0 (f is concave up), and when + 4/3 < x we
have f 00 (x) < 0 (f is concave down).
Evaluating f 00 (x) at the critical value x = 0 gives f 00 (0) > 0, so by the second derivative
test we conclude that f (x) has a local minimum at x = 0. This is the only local extremum.
In summary,
p the graph
p of f (x) haspa local minimum atp(0, f (0)) =
p(0, −1). It has
p inflection
points at (− 4/3, f (− 4/3)) = (− 4/3, −1/2) and (+ 4/3, f (+ 4/3)) = (+ 4/3, −1/2).
Here is a sketch showing the graph of f (x), the extrema, and the inflection points, made by
my computer:
It might help your sketch to observe that limx→∞ f (x) = limx→−∞ f (x) = 1. We say that the
line y = −1 is a horizontal asymptote.
3.5.2. Find two numbers whose difference is 100 and whose product is a minimum.
Call the two numbers x and y. We assume that x − y = 100, and we want to minimize the
“product function” P (x, y) = xy. First observe that we can use the condition x − y = 100 to
express P (x, y) as a function of x alone:
P (x) = xy = x(x − 100) = x2 − 100x.
To minize P (x) we first compute the derivative P 0 (x) = 2x − 100. Setting this equal to zero
gives 2x − 100 = 0, hence x = 50. Is this a minimum or a maximum? The second derivative
is P 00 (x) = 2. Since P 00 (50) = 2 > 0, the second derivative test says that P (x) has a minimum
when x = 50. The other number is y = x − 100 = 50 − 100 = −50.
The minimum possible value of the product function is P (x) is P (50) = −2500. Here is the
graph of P (x):
3.5.4. The sum of two positive numbers is 16. What is the smallest possible value of the sum
of their squares?
Call the two numbers x and y. We assume that x + y = 16, and we want to minimize the
“sum of squares function” S(x, y) = x2 + y 2 . We can use the condition x + y = 16 to express
S(x, y) as a function of x alone:
S(x) = x2 + y 2 = x2 + (16 − x)2 = 2x2 − 32x + 256.
To minimize S(x) we first compute the derivative S 0 (x) = 4x − 32. Setting this equal to zero
gives 4x − 32 = 0, hence x = 8. The second derivative is S 00 (x) = 4. Since S 00 (8) = 4 > 0,
the second derivative test says that S(x) has a minimum when x = 8. The other number is
y = 16 − x = 16 − 8 = 8.
The minimum possible value of the sum of squares function is S(8) = 128. Here is the
graph of S(x):